Gates SA-134/MO-3638 Issues

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Reviewing further, I think it's really more like 20mA that's needed.
That measurement is given for the 7K UTC version which is half the DC resistance of the Triad; by nature the higher DCR will mean less current given the same B+. There's some fudge in that measurement too as it's given in a couple different amps that have a 20V B+ difference also.

Still 33% more than 15mA. You could change the cathode resistor to a higher value that gets current down, which will also take max level down a bit. You could also redesign the feedback to use the tertiary winding on that other transformer, which would make it like the UTC variation SA-20, SA-94. Look at the later M5215(?) remote amp that replaced SA-134 to see the output stage with feedback, that's what this offered transformer is from.
 
Just got off the phone with David from Cinemag. He can make me a transformer with 25mA gap and 4:1 10k:600

I believe this will work in the case that I don't get the proper part.

emmr: Change the negative feedback resistor to lower the max gain. These are way looser on NFB than the console versions.

Im desperately trying to figure this circuit out, my brain hurts.

Cathode resistor would be R14, so could pull that up.

I understand the negative feedback can be used as control voltage to the grid (plate to plate) but I cant see where that would be for the 6SN7 in this circuit.

1641325408047.png
 
almost done with the tear down, 75 EI lam, 1.25" fat stack, , butt stacked but no gap paper, primary wire breaks about every 100 turns, so it has had a hard life, probably on 24/7, even the secondary was cooked, rotten wire , so even if you had gotten the pri leads soldered back, it would have failed down the line or not work at all.

goes sec-pri-sec, 800 turn secondarys, they are connected in parallel, will do a complete wrap up soo,
 
trying to figure out magnetic flux for this transformer,

for that we need volts, but all they give us is 25 dBm, some primitive spec that means nothing to my small mind. i mean i have two brain cells left, and one of them is flickering.

we can google a formula, like

Power = P ref x 10 ^(dBm/10)

Pref is 1milli-watt, or 0.001 watts
dBm is given as 25dBm,

Power=0.001 x 10^(25/10) = 0.001 x 10^2.5 = 0.001 x 316 = 0.316 Watts

power in = power out, so 0.316 watts and 12 K Ohms is what?

brute force- Power = I^2 R
0.316 = I^2 x 12,000 , I^2 = 0.316/12,000 = 0.00513 A or about 5 mils primary current

V=IR so Primary Voltage = .005 x 12,000 =
61.6 Volts RMS-Primary

so secondary volts is what? divide by turns ratio which is 4.5, 61.6/4.5=
13.7 Volts RMS-Secondary

we now have volts, and we know how many turns of pri and sec wire, so we are set.

extra credit points, what is secondary current? use turns ratio but be careful.
stepdown xfmr, volts go down, current goes up, because power has to be the same,

so 4.5 x 5 ma = 22.5 ma secondary current.



pretty hefty transformer me thinks, 13.7 line voltage? drive a truck. or a long cable.
 
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drag out a text book to get flux formula? oh hell no.

we use a pre fab spread sheet for 75 EI and find out that out of a max 20 KG flux, this guy is loafing along at less than 4 k gauss, flux.png
 
Andy ,can you connect a piece of wire from the power supply B+ that used to feed the transformer red wire to the place were the blue wire went(plates of 6SN7)?

then measure the voltage across the 470 ohm cathode resistor , that will tell us the idling plate current, thanks!

or if somebody has one of these amps, they could do the same thing,

the DCR missing from the transformer will not alter the readings that much if you just jumper the tube plates to the supply as the plate resistance of the tube is high compared to the xfmr DCR.

going to wind this guy on a slightly bigger core, 87 EI, which has a wider tongue by 1/8".
then we will have room for a gauge bigger wire which will mean the xfmr will run quite a bit cooler and therefore last a very long time. And we can drop the turns count down to match the flux which will lower DCR which also cools down the xfmr while maintaining the saturation point of the core to keep the sound the same.
 
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With the transformer primary path jumpered, R14 is reading 6.97 VDC @ 31.4 mA.

1641840016687.png
1641840068088.png

The straight B+ that would hit the transformer is 221 VDC.

This is all with the modified NFB resistor (R11) at 62K. Altough I dont think that would affect this part of the circuit.

Weirdly, when I went to measure the mA of the B+ line, my meter probe would pop/arc. Only tried it twice. Why would that happen? What am I missing? Is my meter not capable of that level of current?
 
check that resistor, s/b 470

7/470 = 15 ma which is more reasonable

i think i guessed 16 on the DC flux worksheet so i need to grow a longer arm to pat myself on the back.
 
Well then we are totally dialed. Doug has guessed 20mA with his napkin math, so he was close too!

How are you calculating the idling plate current from the cathode resistor voltage alone? Is the 31mA reading from that same resistor relevant?
 
With the transformer primary path jumpered, R14 is reading 6.97 VDC @ 31.4 mA.

View attachment 88505
View attachment 88507

The straight B+ that would hit the transformer is 221 VDC.

This is all with the modified NFB resistor (R11) at 62K. Altough I dont think that would affect this part of the circuit.

Weirdly, when I went to measure the mA of the B+ line, my meter probe would pop/arc. Only tried it twice. Why would that happen? What am I missing? Is my meter not capable of that level of current?
When measuring current, you have to put the meter *in series* with the resistor.
You are effectively shorting the cathode to ground, hence the spark!
 
trying to figure out magnetic flux for this transformer,

for that we need volts, but all they give us is 25 dBm, some primitive spec that means nothing to my small mind. i mean i have two brain cells left, and one of them is flickering.

we can google a formula, like

Power = P ref x 10 ^(dBm/10)

Pref is 1milli-watt, or 0.001 watts
dBm is given as 25dBm,

Power=0.001 x 10^(25/10) = 0.001 x 10^2.5 = 0.001 x 316 = 0.316 Watts

power in = power out, so 0.316 watts and 12 K Ohms is what?

brute force- Power = I^2 R
0.316 = I^2 x 12,000 , I^2 = 0.316/12,000 = 0.00513 A or about 5 mils primary current

V=IR so Primary Voltage = .005 x 12,000 =
61.6 Volts RMS-Primary

so secondary volts is what? divide by turns ratio which is 4.5, 61.6/4.5=
13.7 Volts RMS-Secondary

we now have volts, and we know how many turns of pri and sec wire, so we are set.

extra credit points, what is secondary current? use turns ratio but be careful.
stepdown xfmr, volts go down, current goes up, because power has to be the same,

so 4.5 x 5 ma = 22.5 ma secondary current.



pretty hefty transformer me thinks, 13.7 line voltage? drive a truck. or a long cable.

Thank you! This is super helpful, Im understanding this!

Power = P ref x 10 ^(dBm/10)
However, I dont understand this one.
 
most humanoids do not hear things on a linear level. this is to protect your ears against a jet landing in your living room at full throttle with GE JT9D large compression turbofans running on hi octane JP7 fuel. so we hear on a logarathmic (sp) scale.
that formula above simply converts the Log power rating (25 dBm) back to the linear rating (0.316 watts) from which it was derived.

audio Log scales use a reference point, like 0.775 for volts or 1milliwatt for power.

526 dBm is the power generated when a black hole swallows a freight train loaded with pineapples and Samoan shot puts.



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